Find dy/dx were y = (sinx)^cosx
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y = e^log^(sinx)^cosx
y = e^(cosx.logsinx)
dy/dx = e^(cosx.logsinx).d/dx(cosx.logsinx)
y' = e^(cosx.logsinx).(-sinx.logsinx+cosx.<1/sinx>cosx)
y' = e^(cosx.logsinx).(-sinx.logsinx+cosx.tanx)
y' = -(sinx)^cosx.[since.logsinx+cosxtanx]......answer.
anuanu45:
can u answer in a more simple model
how did you get e
I too e to make differentiation easier.
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